[英]Break brick game will only break the bricks in order
我正在做一个破砖游戏。 当球撞到一块砖头时,砖头应该消失了。 问题在于它只会按阵列的顺序打碎砖块。 因此,如果您沿线撞到了第三块砖,但是第一块和第二块砖还没有被打破,那么球只会弹起而砖不会消失。
let xPos = 20;
let yPos = 200;
let xRect = 10;
let yRect = 570;
let xVel = 5;
let yVel = 5;
let xBrick1 = [2];
let xBrick2 = [2];
let xBrick3 = [2];
let yBrick = 0;
let r, g, b, h;
let rS, gS, bS;
function setup() {
createCanvas(500, 600);
strokeWeight(5);
////////////////////////////////////////////
r = Math.round(random(255));
g = Math.round(random(255));
b = Math.round(random(255));
if (r == 0) {
rS = r;
} else {
rS = r-20;
}
if (g == 0) {
gS = g;
} else {
gS = g-20;
}
if (b == 0) {
bS = b;
} else {
bS = b-20;
}
////////////////////////////////////////////////
makeBricks();
//////////////////////////////////////////////////
console.log(r, g, b);
console.log(rS, gS, bS);
}
function draw() {
background(000);
xPos+=xVel;
yPos+=yVel;
xRect = mouseX - 100;
fill("fff");
stroke("#000");
rect(xRect, yRect, 150, 30);
fill("#009900");
stroke("#000");
circle(xPos, yPos, 20);
if (breakBrick() == true) {
yVel *= (-1);
}
//////////////////////////////////////////////////////
for (let i = 0; i < 8; i++) {
fill(r, g, b);
strokeWeight(2);
stroke(rS, gS, bS);
rect(xBrick1[i], 0, 60, 20)
xBrick1[i+1] = xBrick1[i] + 62;
}
for (let q = 0; q < 8; q++) {
fill(r, g, b);
strokeWeight(2);
stroke(rS, gS, bS);
rect(xBrick2[q], 22, 60, 20);
xBrick2[q+1] = xBrick2[q] + 62;
}
for (let s = 0; s < 8; s++) {
fill(r, g, b);
strokeWeight(2);
stroke(rS, gS, bS);
rect(xBrick3[s], 44, 60, 20);
xBrick3[s+1] = xBrick3[s] + 62;
}
/////////////////////////////////////////////////////////
if(xPos > 480 || xPos < 20) {
xVel *= (-1);
}
if(yPos < 20) {
yVel *= (-1);
}
if(hitTest(xPos, yPos, xRect, yRect) == true) {
yVel *= (-1);
}
else if(yPos > 600){
xPos = 250;
yPos = 530;
yVel *= (-1);
}
}
function hitTest(xPos, yPos, xRect, yRect) {
if (xPos +20 > xRect && xPos < xRect + 150) {
if (yPos + 20 > yRect && yPos < yRect + 30) {
return true;
}
}
return false;
}
function makeBricks() {
for (let ip = 0; ip < 8; ip++) {
fill(r, g, b);
strokeWeight(2);
stroke(rS, gS, bS);
xBrick1.push(xBrick1[ip]);
rect(xBrick1[ip], 0, 60, 20);
xBrick1[ip+1] = xBrick1[ip] + 62;
}
for (let qp = 0; qp < 8; qp++) {
fill(r, g, b);
strokeWeight(2);
stroke(rS, gS, bS);
xBrick2.push(xBrick2[qp]);
rect(xBrick2[qp], 22, 60, 20);
xBrick2[qp+1] = xBrick2[qp] + 62;
}
for (let sp = 0; sp < 8; sp++) {
fill(r, g, b);
strokeWeight(2);
stroke(rS, gS, bS);
xBrick3.push(xBrick3[sp]);
rect(xBrick3[sp], 44, 60, 20);
xBrick3[sp+1] = xBrick3[sp] + 62;
}
}
function breakBrick() {
for (let h = 0; h < 8; h++) {
if (xPos +20 > xBrick3[h] + 20 && xPos < xBrick3[h] + 60) {
if (yPos +20 > 44 && yPos < 84) {
xBrick3.splice(h, 1);
return true;
}
}
if (xPos +20 > xBrick2[h] + 20 && xPos < xBrick2[h] + 60) {
if (yPos +20 > 22 && yPos < 62) {
xBrick2.splice(h, 1);
return true;
}
}
if (xPos +20 > xBrick1[h] + 20 && xPos < xBrick1[h] + 60) {
if (yPos +20 > 0 && yPos < 40) {
xBrick1.splice(h, 1);
return true;
}
}
}
}
我已经浏览了所有代码和注释,但无法弄清楚。
我该如何解决这个问题?
如果尝试调试它,您会注意到砖实际上已被销毁,但随后在draw
函数中立即被覆盖。 当我注释掉3行时,游戏似乎运行正常:
for (let i = 0; i < 8; i++) {
fill(r, g, b);
strokeWeight(2);
stroke(rS, gS, bS);
rect(xBrick1[i], 0, 60, 20)
// xBrick1[i+1] = xBrick1[i] + 62;
}
for (let q = 0; q < 8; q++) {
fill(r, g, b);
strokeWeight(2);
stroke(rS, gS, bS);
rect(xBrick2[q], 22, 60, 20);
// xBrick2[q+1] = xBrick2[q] + 62;
}
for (let s = 0; s < 8; s++) {
fill(r, g, b);
strokeWeight(2);
stroke(rS, gS, bS);
rect(xBrick3[s], 44, 60, 20);
// xBrick3[s+1] = xBrick3[s] + 62;
}
我相信您可以自己弄清楚其余的东西,这里的教训是: 您应该避免变异 。 在这里阅读有关为什么这是一种不好的做法。
可预测性
突变会隐藏变化,从而产生(意外的)副作用,这可能会导致讨厌的错误。 当您实施不变性时,可以使您的应用程序体系结构和思维模型保持简单,这使您更容易推理出应用程序。
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